我有两个 go 语言的应用程序。user_management 应用程序，我首先运行 (docker-compose up --build)，然后运行 (docker-compose up --build) sport_app。sport_app 依赖于 user_management 应用程序。
sport_app Dockerfile 文件如下。
FROM golang:alpine
RUN apk update && apk upgrade && apk add --no-cache bash git openssh curl
WORKDIR /go-sports-entities-hierarchy
COPY . /go-sports-entities-hierarchy/
RUN rm -rf /go-sports-entities-hierarchy/.env
RUN go mod download
RUN chmod +x /go-sports-entities-hierarchy/scripts/*
RUN ./scripts/build.sh
ADD https://github.com/ufoscout/docker-compose-wait/releases/download/2.2.1/wait /wait
RUN chmod +x /wait
ENV GIN_MODE="debug" \
GQL_SERVER_HOST="localhost" \
GQL_SERVER_PORT=7777 \
ALLOWED_ORIGINS=* \
USER_MANAGEMENT_SERVER_URL="http://localhost:8080/user/me" \
# GQLGen config
GQL_SERVER_GRAPHQL_PATH="graphql" \
GQL_SERVER_GRAPHQL_PLAYGROUND_ENABLED=true \
GQL_SERVER_GRAPHQL_PLAYGROUND_PATH="playground" \
# …我对简单程序的输出感到困惑。我希望在输出中获得所有四个名称，但无法在输出中获得第一个名称。请帮我清除这个问题，或者一些相关的资源。
type Employees struct {
Name string
}
func main() {
chandran := Employees{Name: "Chandran"}
darpan := Employees{Name: "Darpan"}
ishaan := Employees{Name: "Ishaan"}
manbir := Employees{Name: "Manbir"}
Employees.structName(chandran, darpan, ishaan, manbir)
}
func (e Employees) structName(names ...Employees){
fmt.Println(names)
}